I=∫02πln(1+cosx)dx=2∫0πln (1+cosx)dx[∵∫02af(x)=2∫0af(x)dxIf f(2a−x)=f(x)here Cos(2π−x)=cosx]I=2∫0πln(1+cos(π−x))=2∫0πln(1−cosx)dx2I=2∫0πln(1+cosx)+ln(1−cosx)dx=2∫0πln(1−cos2x)dx=2∫0πln Sin2x dx=4∫0πln sinx dx=(4)(2)∫0π/2ln sinx dx[∴Sin(π−x)=sinx]2I=(8)(−π2log2)=−4πlog2I=−2π log2[∴∫0π/2ln sinxdx=−π2log2]

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$\int_{0}^{2 π} \ln (1 + C o s x) d x =$

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$- π \ln 2$

$- 2 π \ln 2$

$2 π \ln 2$

$π \ln 2$

answer is C.

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Detailed Solution

$\begin{array}{l} I = \int_{0}^{2 π} \ln (1 + \cos x) d x \\ = 2 \int_{0}^{π} \ln (1 + \cos x) d x \\ [∵ \int_{0}^{2 a} f (x) = 2 \int_{0}^{a} f (x) d x \\ I f f (2 a - x) = f (x) \\ h e r e C o s (2 π - x) = \cos x] \\ I = 2 \int_{0}^{π} \ln (1 + \cos (π - x)) = 2 \int_{0}^{π} \ln (1 - \cos x) d x \\ 2 I = 2 \int_{0}^{π} \ln (1 + \cos x) + \ln (1 - \cos x) d x \\ = 2 \int_{0}^{π} \ln (1 - \cos^{2} x) d x \\ = 2 \int_{0}^{π} \ln S i n^{2} x d x \\ = 4 \int_{0}^{π} \ln \sin x d x \\ = (4) (2) \int_{0}^{π / 2} \ln \sin x d x \\ [∴ S i n (π - x) = \sin x] \\ 2 I = (8) (- \frac{π}{2} \log 2) = - 4 π \log 2 \\ I = - 2 π \log 2 \\ [∴ \int_{0}^{π / 2} \ln \sin x d x = \frac{- π}{2} \log 2] \end{array}$